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Question
Three fair coins are tossed. What is the probability of getting three heads given that at least two coins show heads?
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Solution
When three fair coins are tossed, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
∴ n(S) = 8
Let event A: Getting three heads.
∴ A = {HHH}
Let event B: Getting at least two heads.
∴ B = {HHT, HTH, THH, HHH}
∴ n(B) = 4
∴ P(B) = `("n"("B"))/("n"("S")) = 4/8`
Now, A ∩ B = {HHH}
∴ n(A ∩ B) = 1
∴ P(A ∩ B) = `("n"("A" ∩ "B"))/("n"("S")) = 1/8`
∴ Probability of getting three heads, given that at least two coins show heads, is given by
`"P"("A"/"B") = ("P"("A" ∩ "B"))/("P"("B")`
= `(1/8)/(4/8)`
= `1/4`
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